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EJ2210 - Analysis of Electrical Machines
FEM-Based Analysis of Induction Machines
Oskar WallmarkAssociate Professor, PhD
Laboratory of Electrical Energy Conversion
Royal Institute of Technology
Stockholm, Sweden
May 4, 2012
◮ Many of you have previous experience with FEM.
◮ Which of Maxwell’s equations are solved for in thisapplication?
◮ How are magnetic field lines determined?
◮ How is the flux linkage (and voltage) in the coils computed?
◮ How can torque be computed?
◮ Why is the mesh density so high in the air gap?
◮ Isn’t the global accuracy of the obtained solution disturbed ifthe mesh density is high only in the air gap?
Review of Maxwell’s Equations
Basic quantities:
◮ Electric field strength E=Ex x + Ey y + Ez z [V/m].
◮ Magnetic field strength H [A/m].
◮ Electric flux density D [C/m2].
◮ Magnetic flux density B [Vs/m2], [T].
◮ Current density J [A/m2].
◮ Electric charge density ρ [C/m3].
◮ Note that E, H, D, B and J are vector fields and ρ is a scalarfield.
Review of Maxwell’s Equations
And X 1 said:
∇× E = −∂B
∂t(Faraday’s law)
∇×H = J+∂D
∂t≈ J (Ampere’s law, quasi static form)
∇ ·D = ρ (Gauss’ law for electric fields)
∇ ·B = 0 (Gauss’ law for magnetic fields).
◮ The current density represented by ∂D/∂t can be neglectedbelow radio frequencies (D=εE and ε is “very small,”ε0=8.85 · 10−12 As/Vm).
◮ Four coupled partial differential equations (PDEs).
◮ For our application, is it possible to reduce the above to onePDE?
1Insert suitable object of worship.
Magnetic Vector Potential
◮ Introduce the magnetic vector potential defined as B=∇×A.
◮ A is not uniquely defined from B=∇× A, but is if we alsodefine ∇ ·A. We choose ∇ ·A=0 for simplicity.
◮ From Faraday’s law, we have (changed order of ∇× and ∂/∂t)
∇× E = −∇×∂A
∂t. (1)
◮ From (1), we have (since ∇×∇V ≡0)
E = −∂A
∂t−∇V (2)
where V is the reduced electric scalar potential.
Magnetic Vector Potential
◮ Ohm’s law tells us that J=σE, where σ is the electrical
conductivity [S/m].
◮ B and H are related as B=µH, where µ is the permeability
[Vs/Am].
◮ Express Ampere’s law (quasi static) using A yields
∇×
(1
µ∇× A
)
= J (3)
∇×
(1
µ∇× A
)
+ σ∂A
∂t+ σ∇V = 0 (4)
where (3) is solved in regions where J is defined (stator slots)and areas “without” current densities (laminations) and (4) issolved in regions where eddy currents may be induced (rotorbars).
Magnetic Vector Potential
◮ In an electric machine, axial effects can often be neglected,i.e, B=Bx x + By y and J=Jz z is now assumed (a twodimensional problem).
◮ Since ∇×
(1
µ∇× A
)
=J and J=Jz z ⇒ A=Az z.
◮ To summarize: B has only x- and y -components and J and A
have only a z-component.
Magnetic Vector Potential
◮ How is the term σ∇V in (4) determined?
◮ The two first components on the right-hand side of (4)contain only z-components. Hence, ∇V contains only az-component, i.e.,
∇V =∂V
∂xx +
∂V
∂yy +
∂V
∂zz =
∂V
∂zz.
V could be a linear function of z (V =C0 + C1z), but sincethis only yields an additional constant ∇V =C1z, V ≡0 ischosen in our application.
◮ A non-zero V can be used to force constraints on the currentin conductive regions. Thereby, a connection to a circuit withvoltage sources as input can be realized.
Problem Formulation
◮ Let’s review the equations to solve again (two dimensionalcase)
∇×
(1
µ∇× Az z
)
= Jz z (5)
∇×
(1
µ∇× Az z
)
+ σ∂Az z
∂t= 0. (6)
◮ Eqs. (5) and (6) are solved in FEM-based softwares such asFlux, JMAG, Opera, Comsol (including three-dimensionalformulations).
◮ Note that only (5) has to be solved if eddy currents areneglected ⇒ A PM motor can be analyzed using only (5).
Drawing Flux Lines
◮ The FEM solver solves for A and B is computed fromB=∇×A. In the two-dimensional case
B = Bx x + By y = ∇× Az z =∂Az
∂yx −
∂Az
∂xy .
◮ Flux lines are lines to which the flux density is parallel.
◮ If Az is constant along the x-direction ⇒ Bx=∂Az/∂y andBy =−∂Az/∂x = 0.
◮ If Az is constant along the y -direction ⇒ Bx =∂Az/∂y=0and By =−∂Az/∂x .
◮ Same argument holds for any direction in which Az is constant⇒ Flux lines can be defined as lines on which Az is constant.
Boundary Conditions
◮ Proper boundary conditions need to be added to solve (5) and(6).
◮ Setting Az=0 on the outer stator periphery will force the fluxlines to be parallel along this boundary (i.e., the flux density isconfined within the machine).
◮ This boundary condition is called a homogenous Dirichletboundary condition.
◮ If a complete machine is simulated, the boundary conditionabove is sufficient.
Boundary Conditions
◮ If only a portion of the machine is simulated, proper boundaryconditions need to be added.
◮ The (homogenous) Neumann boundary condition
∂Az
∂n= 0
forces the flux lines to cross the boundary in parallel to thenormal direction.
◮ Antiperiodic boundary conditions such as
Az(r , ϕ) = −Az
(
r , ϕ+(2k − 1)π
p/2
)
, k = 1, 2, 3, . . . , p/2
enable the study of an odd number of poles (including a singlepole).
Computing Flux and Flux Linkage
◮ For an electric machine with active length La, the flux φbetween points (x1, y1) and (x2, y2) can be computed as
φ =
∫
S
B · ndS =
∫
S
(∇× A) · ndS
=
∮
C
A · d r = La (Az(x1, y1)− Az(x2, y2)) .
◮ For massive conductors with cross sections SCu, the flux canbe computed using the average value Az on the conductorsurface
φ =La
SCu
(∫
S+Cu
AzdS −
∫
S−
Cu
AzdS
)
.
◮ Flux linkage ψ simply obtained by multiplying φ by thenumber of turns.
◮ Induced voltage v=dψ/dt.
Computing Torque
◮ How can torque be computed?
◮ Let’s start with the well known Lorenz force law (force oncharge q traveling with the velocity v
F = q (E+ (v × B)) .
◮ Lorenz force law is an empirical statement and can be used todefine E and B.
◮ Lorenz force law in differential form, assuming vacuum µ=µ0and neglecting the contribution from E
dF = fdV = J× BdV =
(1
µ0∇× B
)
× B
︸ ︷︷ ︸
= f
dV .
Computing Torque
◮ Hence, f=
(1
µ0∇× B
)
× B. The x-component of f can be
expressed as (assuming Bz=0)
fx =1
µ0
(
−By∂By
∂x+ By
∂Bx
∂y
)
.
◮ Add and subtract the term (1/µ0)Bx∂Bx/∂x and use theidentity ∂(B2
x )/∂x=2Bx∂Bx/∂x yields
fx =1
µ0
1
2
∂
∂x
(B2x
)+ By
∂Bx
∂y−
1
2
∂
∂x(B2
x + B2y
︸ ︷︷ ︸
= B2
)
.
Computing Torque
◮ Further manipulation yields
fx =1
µ0
∂
∂x
(
B2x −
B2
2
)
+∂(BxBy )
∂y− Bx ∇ ·B
︸ ︷︷ ︸
= 0
=1
µ0
(∂
∂x
(
B2x −
B2
2
)
+∂(BxBy)
∂y
)
.
◮ Now, fx can be expressed as fx =∇ · Sx where
Sx =1
µ0
((B2x − B2/2
)x + BxBy y
).
◮ Similarly, fy can be expressed as fy =∇ · Sy where
Sy =1
µ0
(BxBy x +
(B2y − B2/2
)y).
Computing Torque
◮ We have
f = fx x + fy y = ∇ · Sx x +∇ · Sy y =
[∇ · Sx
∇ · Sy
]
. (7)
◮ Eq. (7) can be expressed as
f = ∇ · S where S =1
µ0
[B2x − B2/2 BxBy
ByBx B2y − B2/2
]
. (8)
◮ The matrix S is called Maxwell’s stress tensor (assuming atwo-dimensional geometry and without electrical fields).
◮ Note that ∇ · S is a vector.
Computing Torque
◮ What is the point of deriving the expression f = ∇ · S?
◮ Answer: Because we can now use the divergence theorem
F =
∫
df =
∫
V
fdV =
∫
V
∇ · SdV =
∮
S
S · ds. (9)
◮ From (9), we have
df=S · ds. (10)
◮ From (10), the differential torque dT can be expressed as
dT = rdfϕ (11)
where dfϕ is the tangential component of df, i.e., dfϕ=df · ϕ.
Computing Torque
◮ Let’s find dT expressed using cylindrical coordinatesB=Br r + Bϕϕ
dT = rdfϕ = rdf · ϕ = r (S · ds) · ϕ. (12)
◮ We have ds=Lardϕr=Lardϕ(sinϕx + cosϕy). Also,ϕ=− sinϕx + cosϕy .
◮ We can also express Bx and By (in the matrix S) in terms ofBr and Bϕ according to
Bx = Br cosϕ− Bϕ sinϕ
By = Br sinϕ+ Bϕ cosϕ.
◮ Insert the expressions above in (12) yields
T =
∫
dT = Phew! =Lar
2
µ0
∫ 2π
0
BrBϕdϕ.
Computing Torque
◮ Let’s recapitulate:
T =Lar
2
µ0
∫ 2π
0
BrBϕdϕ.
◮ The integral should be evaluated in the air gap.
◮ Both Br and Bϕ are required in the computation.
◮ Br and Bϕ are obtained from B=∇× A.
◮ A is solved for in each node. Between the nodes, linearapproximations are assumed.
◮ Hence, to calculate derivatives of A (to obtain components ofB), a suitably high mesh density is needed.
◮ Since the torque contains a product of two components of B,the mesh density in the air gap needs to be even higher.
◮ Other methods for torque computation are also available(virtual work and the flux times current relationship).
Can we now answer the initial questions?
◮ Which of Maxwell’s equations are solved for in thisapplication?-Answer: We combined them and derived a single equation tobe solved involving only Az (the magnetic vector potential).
◮ How are magnetic field lines determined?-Answer: Magnetic field lines are simply contours where Az isconstant.
◮ How is the flux linkage (and voltage) in the coils computed?-Answer: The flux can be obtained simply from the differenceof Az between the two points.
◮ How can torque be computed?-Answer: A common method was presented where the torquewas found by evaluating a derived line integral (obtained fromMaxwell’s stress tensor) in the air gap.
Can we now answer the initial questions?
◮ Why is the mesh density so high in the air gap?-Answer: A product of two components of B is required in thetorque computation.
◮ Isn’t the global accuracy of the obtained solution disturbed ifthe mesh density is high only in the air gap?-Answer: Globally, the mesh density only needs to be so highso that Az can be resolved with sufficient accuracy.
Introduction to FEMM
◮ FEMM is a free finite element based software.
◮ FEMM can be downloaded from www.femm.info
◮ FEMM solves two-dimensional (planar and axial symmetry)magnetostatic, time-harmonic magnetic, electrostatic andstatic heat flow problems.
Introduction to FEMM
◮ We need to solve
∇×
(1
µ∇× Az z
)
= Jz z (13)
∇×
(1
µ∇× Az z
)
+ σ∂Az z
∂t= 0. (14)
◮ FEMM can solve (13) but not (14) in its present form.
Introduction to FEMM
◮ Assume that all fields are varying sinusoidally with a single,fixed frequency ω.
◮ Introduce the jω-phasor notation
A = Az(x , y)z ejωt .
◮ Insert A into (14) yields
∇×
(1
µ∇× Az z
)
+ jωσAz z = 0. (15)
◮ FEMM can solve (15). To obtain Az =Az(x , y , t), we compute
Az = ℜ(Azejωt) = ℜ(Az) cos ωt −ℑ(Az) sinωt. (16)
◮ Note that the output of FEMM (Az , Bx , By , ...) are hencecomplex quantities and time dependence is obtained byapplying (16).
Introduction to FEMM
◮ Caution: By considering only a single stator and rotorfrequency, effects due to harmonics are neglected. This cancause large errors in the predicted torque for large slip values.
◮ For the phasor notation to be mathematically correct, alloperators need to be linear. Therefore, the permeability µshould be constant (magnetic saturation cannot be modeledexactly).
◮ Saturation can be modeled approximately by inserting anequivalent permeability as outlined in, e.g.,N. Bianchi, Electrical Machine Analysis Using Finite Elements,CRC Taylor & Francis, 2005.
◮ Note that if saturation is neglected, care must be taken whenmodeling machines with closed rotor slots.
A Very Brief Introduction to FEM
◮ The equations to solve can be expressed as
Lφ = f .
◮ φ is the unknown function (Az in our case). Fortwo-dimensional problems φ=φ(x , y).
◮ L is the differential operator (=∇×1
µ∇× in our case).
◮ f is the forcing function (=Jz in our case).
Galerkin’s Method
◮ Introduce φ∗ that approximates the exact solution φ.
◮ Now, φ∗ is expressed as
φ∗ =N∑
j=1
Φjνj
where Φj are unknown coefficients (constants) that must bedetermined and νj is a defined set of base functions.
◮ Introduce the residual r=Lφ∗ − f which ideally should bezero.
Galerkin’s Method
◮ Instead of demanding r=0, let’s relax that condition and letthe integral (volume integral over the problem domain V ) ofthe residuals ri , weighted with some weight function wi , bezero instead. Hence,
ri =
∫
V
wi (Lφ∗− f ) dV = 0.
◮ Now, select wi =vi (known as Galerkin’s method) yields
ri =
∫
V
νiL
N∑
j=1
Φjνj
− νi fdV = 0, i = 1, 2, 3, . . . ,N
(17)
Galerkin’s Method
◮ Eq. (17) can be expressed on matrix form as
SΦ = T (18)
where
Sij =
∫
V
νiLνjdV
Φ = [Φ1 Φ2 Φ3 · · · ΦN ]T
Tij =
∫
V
νi fdV .
Galerkin’s Method
◮ We have transformed the original problem (the PDE) to (18)which is a large matrix equation with the unknowns Φi .
◮ The resulting large matrix S is often sparse and symmetrical.
◮ Much research has been focused on developing methods tosolve (18) with high computational efficiency given that S issparse and symmetrical.
◮ Other approaches also exist that transforms the originalproblem into a large matrix equation (e.g. the Rayleigh-Ritz’svariational approach).
The Finite Element Method
◮ Note that we have not defined the base functions νj yet.
◮ In the finite element approach, the problem domain is typicallydivided into a number of triangles (two-dimensional case).
◮ The (unkown) solution in element (triangle) m can beexpressed as
φ∗m(x , y) =
3∑
j=1
Φmjνmj
where Φmj is the value of φ in the jth node of the mthelement.
The Finite Element Method
◮ In the finite element method, the base functions νmj arechosen so that
νmj(x , y) =
{1, in node j
0, in all other nodes.
◮ Between all nodes, linear interpolation is used.
◮ The resulting matrix S becomes sparse and the non-zeroelements are simple to compute ⇒ computational efficiency.
Modeling Non-Linear Materials
◮ In electrical steels, the permeability µ is strongly nonlinear.Hence, µ=µ(B).
◮ We have seen how the finite element approach results in alarge matrix equation.
◮ The nonlinearity is typically solved using the Newton-Raphsonapproach.
◮ As an example, for the nonlinear system F(x)=0, the solutionx is found by the iteration
xk+1 = xk −
(∂F
∂x
)−1
F(xk)
where the partial derivatives are evaluated at x=xk .
◮ The approach above is implemented and the iterationcontinues until the change is smaller than a certain threshold.
What have we learnt?
◮ We have seen how the original PDE can be transformed into alarge matrix equation.
◮ The resulting solution is an approximation of the exactsolution.
◮ We have seen how the finite element approach results in amatrix S that is sparse and where the non-zero elements aresimple to compute.
◮ Much more can be said about the finite element method butwhat mentioned is sufficient for our purposes.